Answer:
the energy absorbed is 4.477 x 10⁶ J
Explanation:
mass of the liquid, m = 13 kg
initial temperature of the liquid, t₁ = 18 ⁰C
final temperature of the liquid, t₂ = 100 ⁰C
specific heat capacity of water, c = 4,200 J/kg⁰C
The energy absorbed is calculated as;
H = mcΔt
H = mc(t₂ - t₁)
H = 13 x 4,200(100 - 18)
H = 4.477 x 10⁶ J
Therefore, the energy absorbed is 4.477 x 10⁶ J
